The index of refraction of a transparent liquid
(similar to water but with a different index of refraction) is 1.39. A flashlight held under the
transparent liquid shines out of the transparent liquid in a swimming pool. This beam of light exiting the surface of the transparent liquid makes an angle of �a = 39 degrees with respect
to the vertical.
At what angle (with respect to the vertical)is the flashlight being held under transparent liquid?
Answer in units of degrees.
The flashlight is slowly turned away from the vertical direction.
At what angle will the beam no longer be visible above the surface of the pool? Answer in units of degrees.
We can use Snell's Law: n1 * sin(theta1) = n2 * sin(theta2), where n1 and n2 are indices of refraction, and theta1 and theta2 are the incident and refracted angles, respectively.
For the first part:
n1 = 1.39 (from the problem statement)
n2 = 1 (assuming the air's index of refraction)
theta2 = 39 degrees
Solving for theta1:
1.39 * sin(theta1) = 1 * sin(39) = 0.629
theta1 = arcsin(0.629 / 1.39) = 27.35 degrees (approximately)
So the flashlight is being held at approximately 27.35 degrees with respect to the vertical under the transparent liquid.
For the second part, we want to find the critical angle, the angle at which light will no longer be visible above the surface of the pool. This occurs when the light exits at an angle of 90 degrees with respect to the vertical. So:
n1 * sin(theta1_critical) = n2 * sin(90)
1.39 * sin(theta1_critical) = 1
theta1_critical = arcsin(1 / 1.39) = 46.91 degrees (approximately)
So the beam of light will no longer be visible above the surface of the pool if the flashlight is held at an angle greater than 46.91 degrees (approximately) with respect to the vertical.
To find the angle at which the flashlight is being held under the transparent liquid, we can use Snell's Law, which relates the angle of incidence to the angle of refraction when light passes through the boundary between two mediums.
The equation for Snell's Law is: n1 * sin(θ1) = n2 * sin(θ2)
Where:
n1 = index of refraction of the first medium (in this case, air) = 1 (approximate)
θ1 = angle of incidence (unknown)
n2 = index of refraction of the second medium (transparent liquid) = 1.39
θ2 = angle of refraction (39 degrees)
We can rearrange the equation to solve for θ1:
sin(θ1) = (n2 / n1) * sin(θ2)
sin(θ1) = (1.39 / 1) * sin(39)
sin(θ1) ≈ 1.39 * 0.63
sin(θ1) ≈ 0.876
Now, we can take the inverse sine (sin^-1) of both sides to find θ1:
θ1 ≈ sin^-1(0.876) ≈ 60.45 degrees
Therefore, the flashlight is being held at an angle of approximately 60.45 degrees with respect to the vertical under the transparent liquid.
To find the angle at which the beam will no longer be visible above the surface of the pool, we can use the concept of total internal reflection. Total internal reflection occurs when the angle of incidence is greater than the critical angle, resulting in all of the light being reflected back into the medium.
The critical angle (θc) can be found using the equation: sin(θc) = n2 / n1
sin(θc) = 1.39 / 1
sin(θc) = 1.39
θc ≈ sin^-1(1.39) ≈ 53.35 degrees
Since the angle of refraction exceeds the critical angle at this angle, the beam will no longer be visible above the surface of the pool.
Therefore, the angle at which the beam will no longer be visible above the surface of the pool is approximately 53.35 degrees.
To find the angle at which the flashlight is being held under the transparent liquid, we can use Snell's law, which relates the angle of incidence and the angle of refraction when light passes from one medium to another.
Snell's law is given by: n₁sin(θ₁) = n₂sin(θ₂)
Where:
- n₁ is the index of refraction of the medium through which the incident light is traveling (in this case, the transparent liquid)
- θ₁ is the angle of incidence (between the incident ray and the normal line)
- n₂ is the index of refraction of the medium into which the light is entering (in this case, the air)
- θ₂ is the angle of refraction (between the refracted ray and the normal line)
In this problem:
- n₁ = 1.39 (index of refraction of the transparent liquid)
- θ₂ = 39 degrees (angle made by the exiting beam with respect to the vertical)
Let's find θ₁, the angle of incidence:
n₁sin(θ₁) = n₂sin(θ₂)
1.39sin(θ₁) = 1sin(39)
Now we can solve for θ₁:
θ₁ = sin⁻¹(sin(39) / 1.39)
θ₁ ≈ 16.5 degrees
Therefore, the flashlight is being held at an angle of approximately 16.5 degrees with respect to the vertical direction.
To find the angle at which the beam will no longer be visible above the surface of the pool, we need to determine the critical angle. The critical angle occurs when the angle of refraction becomes 90 degrees.
Using Snell's law again, we can solve for the critical angle:
n₁sin(θ_c) = n₂sin(90)
1.39sin(θ_c) = 1sin(90)
θ_c = sin⁻¹(1 / 1.39)
θ_c ≈ 48.3 degrees
Therefore, when the flashlight is turned beyond an angle of approximately 48.3 degrees from the vertical, the beam will no longer be visible above the surface of the pool.